Separation of variables in Fourier Series

• Sep 15th 2007, 04:25 PM
Separation of variables in Fourier Series
Problem: Find the Fourier Series of $\displaystyle f(x) = 0.05sin \pi x$, with the initial velocity $\displaystyle g(x)=0$, $\displaystyle c= \frac{1}{n}$, $\displaystyle L = 1$.

My solution: I use the separation of variables method, then I have:

$\displaystyle u(x,t)= \sum^{\infty}_{n=1}sin \frac{n \pi x}{L}[b_{n}cos( \frac {cn \pi}{L})t+ \hat{b_{n}} sin( \frac {cn \pi}{L})t]$

where

$\displaystyle b_{n}=\frac{2}{L} \int^{L}_{0}f(x)sin( \frac {cn \pi}{L})dx$
$\displaystyle \hat{b_{n}} = \frac{2}{cnx} \int^{L}_{0}g(x)sin( \frac {cn \pi}{L})dx$

Now I know that [tex]\hat{b_{n}}=0[tex] since $\displaystyle g(x)=0$, but when I solve for $\displaystyle b_{n}$, I have:

$\displaystyle b_{n}=2 \int^{1}_{0}0.05sin( \pi x)sin(n \pi x)dx$

I use the table of integrals, and I get bn = 0, is there something I'm doing wrong here? (Most likely)

K
• Sep 15th 2007, 04:50 PM
ThePerfectHacker
What are you trying to do? :confused:
• Sep 15th 2007, 06:37 PM
$\displaystyle \int_0^L \sin \frac{\pi n x}{L}\sin \frac{\pi m x}{L} dx= L\delta_{nm}$
$\displaystyle \int_0^L \sin \frac{\pi nx}{L}\cos \frac{\pi mx}{L}dx=0$
$\displaystyle \int_0^L \cos \frac{\pi nx}{L}\cos \frac{\pi mx}{L} dx = L\delta_{nm}$